CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-05

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Path-connected spaces have zero reduced zero-th homology

Statement

If $X$ is nonempty and path-connected, then ${\tilde{H}}_{0}^{s i n g} (X; Z) = 0.$

Facts & Assumptions

Given: A nonempty path-connected topological space $X$ .

[L1]

$H_{0}^{s i n g} (X; Z)$ is the free abelian group on the path components of $X$ (Zero-th singular homology is free on path components).

[L2]

Reduced degree-zero singular homology is the homology of the augmentation kernel in degree $0$ (Augmentation at 0-simplices and reduced singular homology).

[L3]

Every singular $1$ -boundary lies in the augmentation kernel (The singular augmentation commutes with the boundary).

Proof

technique · direct

1.1

Since $X$ is nonempty and path-connected, it has exactly one path component. By [L1], $H_{0}^{s i n g} (X; Z) ≅ Z .$

L1given

2.1

Let $z = \sum_{k = 1}^{m} a_{k} σ_{k} \in C_{0} (X; Z)$ . Under the isomorphism of step 1.1, the class $[z]$ maps to $\sum_{k = 1}^{m} a_{k} \in Z$ , which is exactly $ε_{X} (z)$ . Therefore $z \in \ker ε_{X}$ if and only if $[z] = 0$ in $H_{0}^{s i n g} (X; Z)$ , equivalently if and only if $z \in B_{0} (X; Z)$ . By [L3], every $0$ -boundary lies in $\ker ε_{X}$ , so $\ker ε_{X} = B_{0} (X; Z) .$

L3step 1.1algebra

3.1

By [L2], reduced degree-zero homology is ${\tilde{H}}_{0}^{s i n g} (X; Z) = \ker ε_{X} / im \partial_{1} = \ker ε_{X} / B_{0} (X; Z) = 0.$

L2step 2.1∎

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Sources

Allen Hatcher, Algebraic Topology (standard reference, not scraped)
J. Peter May, A Concise Course in Algebraic Topology (standard reference, not scraped)